3.69.25 $\int \frac{4+37x+32x^2+(-1-17x-16x^2)\log (1+x)+(2x+2x^2){\log }^2(1+x)}{16+16x+(-8-8x)\log (1+x)+(1+x){\log }^2(1+x)}dx$

Optimal. Leaf size=16 $$18+x^2-\frac{x}{-4+\log (1+x)}$$

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Rubi [A]  time = 0.41, antiderivative size = 26, normalized size of antiderivative = 1.62, number of steps used = 14, number of rules used = 10, integrand size = 69, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.145, Rules used = {6741, 6742, 2389, 2299, 2178, 2411, 2353, 2297, 2302, 30} $$\begin{array}{c}{x}^2+\frac{x+1}{4-\log (x+1)}+\frac{1}{\log (x+1)-4}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(4 + 37*x + 32*x^2 + (-1 - 17*x - 16*x^2)*Log[1 + x] + (2*x + 2*x^2)*Log[1 + x]^2)/(16 + 16*x + (-8 - 8*x)
*Log[1 + x] + (1 + x)*Log[1 + x]^2),x]

[Out]

x^2 + (1 + x)/(4 - Log[1 + x]) + (-4 + Log[1 + x])^(-1)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4+37x+32x^2+(-1-17x-16x^2)\log (1+x)+(2x+2x^2){\log }^2(1+x)}{(1+x)(4-\log (1+x){)}^2}dx\\ & =\int (2x+\frac{1}{4-\log (1+x)}+\frac{x}{(1+x)(-4+\log (1+x){)}^2})dx\\ & =x^2+\int \frac{1}{4-\log (1+x)}dx+\int \frac{x}{(1+x)(-4+\log (1+x){)}^2}dx\\ & =x^2+\mathrm{Subst}(\int \frac{1}{4-\log (x)}dx,x,1+x)+\mathrm{Subst}(\int \frac{-1+x}{x(-4+\log (x){)}^2}dx,x,1+x)\\ & =x^2+\mathrm{Subst}(\int \frac{e^x}{4-x}dx,x,\log (1+x))+\mathrm{Subst}(\int (\frac{1}{(-4+\log (x){)}^2}-\frac{1}{x(-4+\log (x){)}^2})dx,x,1+x)\\ & =x^2-e^4\text{Ei}(-4+\log (1+x))+\mathrm{Subst}(\int \frac{1}{(-4+\log (x){)}^2}dx,x,1+x)-\mathrm{Subst}(\int \frac{1}{x(-4+\log (x){)}^2}dx,x,1+x)\\ & =x^2-e^4\text{Ei}(-4+\log (1+x))+\frac{1+x}{4-\log (1+x)}-\mathrm{Subst}(\int \frac{1}{x^2}dx,x,-4+\log (1+x))+\mathrm{Subst}(\int \frac{1}{-4+\log (x)}dx,x,1+x)\\ & =x^2-e^4\text{Ei}(-4+\log (1+x))+\frac{1+x}{4-\log (1+x)}+\frac{1}{-4+\log (1+x)}+\mathrm{Subst}(\int \frac{e^x}{-4+x}dx,x,\log (1+x))\\ & =x^2+\frac{1+x}{4-\log (1+x)}+\frac{1}{-4+\log (1+x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 16, normalized size = 1.00 $$\begin{array}{c}-1+x^2-\frac{x}{-4+\log (1+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(4 + 37*x + 32*x^2 + (-1 - 17*x - 16*x^2)*Log[1 + x] + (2*x + 2*x^2)*Log[1 + x]^2)/(16 + 16*x + (-8
- 8*x)*Log[1 + x] + (1 + x)*Log[1 + x]^2),x]

[Out]

-1 + x^2 - x/(-4 + Log[1 + x])

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fricas [A]  time = 0.46, size = 26, normalized size = 1.62 $$\begin{array}{c}\frac{x^2\log (x+1)-4x^2-x}{\log (x+1)-4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x)*log(x+1)^2+(-16*x^2-17*x-1)*log(x+1)+32*x^2+37*x+4)/((x+1)*log(x+1)^2+(-8*x-8)*log(x+1)
+16*x+16),x, algorithm="fricas")

[Out]

(x^2*log(x + 1) - 4*x^2 - x)/(log(x + 1) - 4)

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giac [A]  time = 0.17, size = 15, normalized size = 0.94 $$\begin{array}{c}{x}^2-\frac{x}{\log (x+1)-4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x)*log(x+1)^2+(-16*x^2-17*x-1)*log(x+1)+32*x^2+37*x+4)/((x+1)*log(x+1)^2+(-8*x-8)*log(x+1)
+16*x+16),x, algorithm="giac")

[Out]

x^2 - x/(log(x + 1) - 4)

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maple [A]  time = 0.03, size = 16, normalized size = 1.00

method	result	size
risch	$x^2-\frac{x}{\ln (x+1)-4}$	$16$
norman	$\frac{x^2\ln (x+1)-x-4x^2}{\ln (x+1)-4}$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+2*x)*ln(x+1)^2+(-16*x^2-17*x-1)*ln(x+1)+32*x^2+37*x+4)/((x+1)*ln(x+1)^2+(-8*x-8)*ln(x+1)+16*x+16),
x,method=_RETURNVERBOSE)

[Out]

x^2-x/(ln(x+1)-4)

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maxima [B]  time = 0.39, size = 38, normalized size = 2.38 $$\begin{array}{c}\frac{x^2\log (x+1)-4x^2-x+4}{\log (x+1)-4}-\frac{4}{\log (x+1)-4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x)*log(x+1)^2+(-16*x^2-17*x-1)*log(x+1)+32*x^2+37*x+4)/((x+1)*log(x+1)^2+(-8*x-8)*log(x+1)
+16*x+16),x, algorithm="maxima")

[Out]

(x^2*log(x + 1) - 4*x^2 - x + 4)/(log(x + 1) - 4) - 4/(log(x + 1) - 4)

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mupad [B]  time = 4.27, size = 15, normalized size = 0.94 $$\begin{array}{c}{x}^2-\frac{x}{\ln (x+1)-4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((37*x - log(x + 1)*(17*x + 16*x^2 + 1) + log(x + 1)^2*(2*x + 2*x^2) + 32*x^2 + 4)/(16*x - log(x + 1)*(8*x
+ 8) + log(x + 1)^2*(x + 1) + 16),x)

[Out]

x^2 - x/(log(x + 1) - 4)

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sympy [A]  time = 0.13, size = 10, normalized size = 0.62 $$\begin{array}{c}{x}^2-\frac{x}{\log (x+1)-4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+2*x)*ln(x+1)**2+(-16*x**2-17*x-1)*ln(x+1)+32*x**2+37*x+4)/((x+1)*ln(x+1)**2+(-8*x-8)*ln(x+1
)+16*x+16),x)

[Out]

x**2 - x/(log(x + 1) - 4)

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